OCR C4 2014 June — Question 4 5 marks

Exam BoardOCR
ModuleC4 (Core Mathematics 4)
Year2014
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicAddition & Double Angle Formulae
TypeIntegrate using double angle
DifficultyStandard +0.8 This question requires recognizing that the numerator is cos(2x) and denominator is 1+sin(2x), then using substitution u=1+sin(2x) to integrate. While the double angle identities are standard C4 content, spotting the structure and executing the substitution with correct limits requires solid technique and algebraic fluency beyond routine exercises.
Spec1.05j Trigonometric identities: tan=sin/cos and sin^2+cos^2=11.05l Double angle formulae: and compound angle formulae1.08h Integration by substitution

4 Show that \(\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 - 2 \sin ^ { 2 } x } { 1 + 2 \sin x \cos x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2\).

Question 4:
AnswerMarks Guidance
AnswerMarks Guidance
\(\int\frac{\cos2x}{1+\sin2x}(dx)\)B1* \(\cos2x=1-2\sin^2x\) or \((1+)\sin2x=(1+)2\sin x\cos x\) seen; if B0B0M0A0, SC4 for \(F[x]=\frac{1}{2}\ln(1+2\sin x\cos x)\) or \(\frac{1}{2}\ln(1+\sin2x)\); final mark may still be awarded
B1*Numerator and denominator both correct in the integral soi
\(F[x]=k\ln(1+\sin2x)\) soiM1dep* or \(k\ln(1+u)\) or \(k\ln(u)\) following their substitution www
\(k=\frac{1}{2}\)A1 Correct \(k\) for their substitution
\(\frac{1}{2}\ln(1+\sin(\frac{\pi}{2}))-\frac{1}{2}\ln(1+0)=\frac{1}{2}\ln2\)A1 AG Correct use of limits www; minimum working: \(\frac{1}{2}\ln2-\frac{1}{2}\ln1\) or \(\frac{1}{2}\ln(1+1)\) oe
[5]
# Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\int\frac{\cos2x}{1+\sin2x}(dx)$ | B1* | $\cos2x=1-2\sin^2x$ or $(1+)\sin2x=(1+)2\sin x\cos x$ seen; if B0B0M0A0, SC4 for $F[x]=\frac{1}{2}\ln(1+2\sin x\cos x)$ or $\frac{1}{2}\ln(1+\sin2x)$; final mark may still be awarded |
| | B1* | Numerator and denominator both correct in the integral soi |
| $F[x]=k\ln(1+\sin2x)$ soi | M1dep* | or $k\ln(1+u)$ or $k\ln(u)$ following their substitution www |
| $k=\frac{1}{2}$ | A1 | Correct $k$ for their substitution |
| $\frac{1}{2}\ln(1+\sin(\frac{\pi}{2}))-\frac{1}{2}\ln(1+0)=\frac{1}{2}\ln2$ | A1 AG | Correct use of limits www; minimum working: $\frac{1}{2}\ln2-\frac{1}{2}\ln1$ or $\frac{1}{2}\ln(1+1)$ oe |
| **[5]** | | |
4 Show that $\int _ { 0 } ^ { \frac { 1 } { 4 } \pi } \frac { 1 - 2 \sin ^ { 2 } x } { 1 + 2 \sin x \cos x } \mathrm {~d} x = \frac { 1 } { 2 } \ln 2$.

\hfill \mbox{\textit{OCR C4 2014 Q4 [5]}}